problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
37. (10-11 grades) How many planes are equidistant from four points that do not lie in the same plane? | 37. Taking these points as the vertices of a tetrahedron, it is easy to establish that only seven planes can be drawn equidistant from its vertices. | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
64. (9th grade) The number 7 is raised to the seventh power, the resulting number is again raised to the seventh power, and so on. This process is repeated 1000 times. What is the last digit of this number? | 64. Considering the powers of the number 7 in sequence, we notice that the last digits of these powers repeat every four, so the number given in the problem ends with the digit 7. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
81. The difference $\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}$ is an integer. Find this number. | Solution. Since $40 \sqrt{2}-57<0$, then $|40 \sqrt{2}-57|=$ $=57-40 \sqrt{2}$. Then
$$
\begin{aligned}
A & =\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}= \\
& =\sqrt{57-40 \sqrt{2}}-\sqrt{57+40 \sqrt{2}}
\end{aligned}
$$
Let $57-40 \sqrt{2}=(a+b \sqrt{2})^{2}$, where $a$ and $b$ are unknown coefficients. Then
$$
5... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
89. Simplify the expression:
a) $\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}$
b) $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ | Solution. a) Let $\sqrt{57-40 \sqrt{2}}-\sqrt{57+40 \sqrt{\overline{2}}}=x ; x<0$, since $\sqrt{57-40 \sqrt{2}}<\sqrt{57+40 \sqrt{2}}$.
Square both sides of the equation:
$$
x^{2}=57-40 \sqrt{2}-2 \sqrt{57^{2}-(40 \sqrt{2})^{2}}+57+40 \sqrt{2}
$$
from which
$$
x^{2}=114-2 \sqrt{49}, x^{2}=100, x=-10
$$
(Solution b... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
90. Prove that $\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}=3$. | Solution. Let the value of the expression on the left side of the equality be denoted by $x$. Reasoning as in the solution of Example 89, b, we obtain the equation $x^{3}-3 x-18=0$, from which $x=3$.
## Solving Equations | 3 | Algebra | proof | Yes | Yes | olympiads | false |
91. Let's solve the equation
$$
(3 x-1) \sqrt{x-4}=17 \sqrt{2}
$$ | Solution. Method 1. The domain of expression (3) is the interval $[4 ;+\infty)$. Squaring both sides of the equation, we obtain the equivalent equation
$$
9 x^{3}-42 x^{2}+25 x-582=0
$$
We find the critical points of the function $f(x)=9 x^{3}-42 x^{2}+25 x-582$:
$$
f^{\prime}(x)=27 x^{2}-84 x+25=0
$$
from which $x... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
31. Cube. Holding a model of a cube in your hand so that it can rotate around its longest axis (i.e., around the line connecting opposite vertices), you can wind black yarn around it without any gaps. The yarn will shade only half of the cube (why?). The same can be done with another axis; there are four in total, and ... | 31. The yarn wound around a cube rotating about one of its axes (Fig. 41) will remain only on those edges that do not have common points with the axis of rotation; the yarn will cover half of each face of the cube, i.e., half of the cube's surface.
Now, let's rotate the cube sequentially around each of the four axes, ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
32. Geodesics. This problem does not require knowledge of mathematics. Let's place a rubber band (so-called "prescription", used in pharmacies for packaging medicines) on a stationary cube in such a way that it holds on the cube and does not cross itself.
The line along which this rubber band will lie is called a geod... | 32. We will prove that through each point on the surface of a cube, there pass four different geodesics, and in total, we have seven families of geodesic lines.
If we assume that the cube is smooth, then a rubber band wrapped around it will be arranged in such a way that the perimeter of the polygon it forms will reac... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
34. Unfolding a Cube. Models of polyhedra are made from flat nets. In a net, faces are adjacent to each other along edges, and the model is constructed by folding the cardboard net along the edges. A regular tetrahedron has two such different nets. How many does a cube have? | 34. All existing nets (a total of 11) are shown in Fig. 49. The first six solutions give nets in which four faces of the cube are arranged in one strip of the net. No other solutions of this type exist. The next four nets are those in which there are three faces in one
. Ship $P$ has spotted ship $Q$, which is sailing in a direction perpendicular to $P Q$, maintaining its course. Ship $P$ is chasing $Q$, always heading directly towards $Q$; the speed of both ships is the same at any moment (but can vary over time). Without calculations, it is clear that $P$ is sailin... | 88. Let $\alpha$ denote the instantaneous angle between the direction $P Q$ and the path of ship $Q$ (Fig. 163), and $v$ - the speed of ships $P$ and $Q$ at that moment. The mutual approach of the ships is influenced by the speed $v$ of ship $P$, directed towards $Q$, and the component $v \cos \alpha$ of the speed of s... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 2. Let's determine in which numeral system the following multiplication was performed: $352 \cdot 31=20152$. | Solution. Let $x$ be the base of the numeral system, then the given equality can be written in the form of an equation
$$
\left(3 x^{2}+5 x+2\right)(3 x+1)=2 x^{4}+x^{2}+5 x+2
$$
By performing the multiplication and combining like terms, we get:
$$
2 x^{4}-9 x^{3}-17 x^{2}-6 x=0
$$
It is clear that $x \neq 0$ and t... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 2. Let's find the sum of the cubes of the roots of the equation
$$
x^{3}+2 x^{2}+x-3=0
$$ | S o l u t i o n. This can be done in various ways, for example, by sequentially calculating the sum of the squares of the roots and then the sum of the cubes of the roots. However, we will use a frequently applied identity in mathematics:
$$
\begin{aligned}
a^{3}+b^{3}+c^{3}-3 a b c= & (a+b+c)\left(a^{2}+b^{2}+c^{2}-a... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Let's find pairs of real numbers $x$ and $y$ that satisfy the equation
$$
x^{2}+4 x \cos x y+4=0
$$ | S o l u t i o n. Several methods can be proposed to solve the given equation.
1st method. Since the equation is quadratic (if we do not consider $x$ under the cosine sign), we can express $x$ in terms of trigonometric functions of the angle $x y$:
$$
x=-2 \cos x y \pm \sqrt{4 \cos ^{2} x y-4}=-2 \cos x y \pm 2 \sqrt{... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 5. Let's find all values of $a$ for which the system
$$
\left\{\begin{array}{l}
2^{b x}+(a+1) b y^{2}=a^{2} \\
(a-1) x^{3}+y^{3}=1
\end{array}\right.
$$
has at least one solution for any value of $b,(a, b, x, y \in \mathbf{R})$. | S o l u t i o n. Suppose there exists some value $a$ for which the system has at least one solution, for example, for $b=0$.
In this case, the system will take the form:
$$
\left\{\begin{array}{l}
1=a^{2} \\
(a-1) x^{3}+y^{3}=1
\end{array}\right.
$$
It is clear that this system is consistent only if $a=1$ or $a=-1$.... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 3. Compare the number $a$ with one, if
$$
a=0.99999^{1.00001} \cdot 1.00001^{0.99999} .
$$ | S o l u t i o n. Let's represent the numbers in the given expression as follows:
$$
0.99999=1-\alpha, \quad 1.00001=1+\alpha
$$
where $\alpha=0.00001$. Then
$$
a=(1-\alpha)^{1+\alpha}(1+\alpha)^{1-\alpha}=\left(1-\alpha^{2}\right)\left(\frac{1-\alpha}{1+\alpha}\right)^{\alpha}
$$
Since $1-\alpha^{2}<1$, then $a<1$
... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
E x a m p l e 1. Let's find the limit of the sum
$$
S_{n}=\frac{3}{4}+\frac{5}{36}+\ldots+\frac{2 n+1}{n^{2}(n+1)^{2}}
$$
as $n \rightarrow \infty$. | S o l u t i o n. First, let's try to simplify $S_{n}$. Since the fraction 56
$\frac{2 k+1}{k^{2}(k+1)^{2}}$ can be represented as the difference $\frac{1}{k^{2}}-\frac{1}{(k+1)^{2}}$, then
$$
\begin{gathered}
\frac{3}{4}=1-\frac{1}{2^{2}} \\
\frac{5}{36}=\frac{1}{2^{2}}-\frac{1}{3^{2}} \\
\frac{2 n+1}{n^{2}(n+1)^{2}}=... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 2. Let $y \neq-1$. We set,
$$
x_{1}=\frac{y-1}{y+1}, \quad x_{2}=\frac{x_{1}-1}{x_{1}+1}, \quad x_{3}=\frac{x_{2}-1}{x_{2}+1}, \ldots
$$
What is $y$ if $x_{1978}=-\frac{1}{3}$? | Solution. Substituting the value of $x_{1}$ into the second equality, after simplifications we get:
$$
x_{2}=-\frac{1}{y}
$$
Further,
$$
x_{3}=\frac{y+1}{1-y}, \quad x_{4}=y, \quad x_{5}=\frac{y-1}{y+1}=x_{1}
$$
Therefore,
$$
x_{5}=x_{1}, \quad x_{6}=x_{2}, \quad x_{7}=x_{3}, \quad x_{8}=x_{4}, \ldots, \quad x_{19... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 2. Let's calculate the sum
$$
a^{2000}+\frac{1}{a^{2000}}
$$
if $a^{2}-a+1=0$. | S o l u t i o n. From the equality $a^{2}-a+1=0$, it follows that:
$$
a-1+\frac{1}{a}=0 \text { or } a+\frac{1}{a}=1
$$
Moreover,
$$
a^{3}+1=(a+1)\left(a^{2}-a+1\right)=0
$$
from which $a^{3}=-1$. Now we have:
$$
\begin{aligned}
& a^{2000}+\frac{1}{a^{2000}}=\left(a^{3}\right)^{666} a^{2}+\frac{1}{\left(a^{3}\righ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Points on a Line. If 10 points are placed at equal intervals on a line, they will occupy a segment of length s, and if 100 points are placed, the segment will have a length S. How many times greater is S than s? | 6. Between ten points there are nine intervals, and between a hundred points - ninety-nine. Therefore, $S$ is greater than $s$ by 11 times. | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Stereometric problem. How many faces does a hexagonal pencil have? | 7. It is important to ask: which pencil? If the pencil has not been sharpened yet, then 8, otherwise there may be variations...
Translating the text as requested, while preserving the original line breaks and formatting. | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12. The Musketeers' Journey. The distance between Athos and Aramis, riding on the road, is 20 leagues. In one hour, Athos travels 4 leagues, and Aramis - 5 leagues. What distance will be between them after an hour? | 12. Well, in which direction was each of the musketeers traveling? The problem statement does not mention this. If they were traveling towards each other, the distance between them would be 11 leagues. In other cases (make a diagram!), the possible answers are: 29 leagues; 19 leagues; 21 leagues. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13. An Arabic Tale. A Flock of Pigeons
$>\Delta \theta \Pi \mathrm{V} \oplus \theta\mathbf{v} \Pi \square \nabla \square \Lambda$
$\oplus \Lambda \nabla \theta \Pi \oplus \mathbf{V V} \varnothing \odot$
ロจ৫ఠ<>VIVOO flew up to a tall tree.
Some of the pigeons perched on the branches, while others settled under the t... | 13. From the condition of the problem, it is clear that the number of pigeons sitting on the branches is two more than those sitting below. Further, it follows from the condition that after one of the pigeons flew up to the branch, the number of pigeons sitting on the branch became twice as many as those sitting on the... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In the government of the country of knights and liars, there are 12 ministers. Some of them are liars, and the rest are knights. Once, at a government meeting, the following opinions were expressed: the first minister said, “There is not a single honest person here,” the second said, “There is no more than one hones... | 4. Note that the number of true statements must match the number of honest people in the government. Further, if a statement from any minister is true, then the statements of each minister who spoke after him are also true. In this case, the only statement that would not lead to a contradiction is: "there are no more t... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Around a round table, eight people are sitting, each of whom is either a knight or a liar. When asked who their neighbors are, each of them answered: “My neighbors are a liar and a knight.” How many of them were liars? How would the answer change if nine people were sitting at the table? | 5. 6. At the table, there is at least one liar. Indeed, if only knights were sitting at the table, each knight's statement "next to me sits a knight and a liar" would be false, which is impossible. 2. The neighbors
=7$. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. What can you buy for a ruble? Nine boxes of matches cost 9 rubles and some kopecks, while ten such boxes cost 11 rubles and some kopecks. How much does one box cost? | 6. Note that a box of matches costs more than $\frac{11}{10}$ rubles, but less than $\frac{10}{9}$ rubles. That is, more than 1.10 rubles, but less than 1.111 rubles. Answer: 1 ruble 11 kopecks. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12. The Big Laundry. After seven hours of washing, the length, width, and height of the soap piece were halved. For how many washes will the remaining soap last? | 12. Since the length, width, and height of the soap piece have been halved, its volume has decreased by 8 times, meaning that in 7 hours, the soap piece has decreased by $\frac{7}{8}$ of its volume (by $\frac{1}{8}$ of its volume per hour). Therefore, the soap will last for one more hour of heavy washing. | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. $о$ about the fisherman and the fish. When asked how big the fish he caught was, the fisherman said: “I think that its tail is 1 kg, the head is as much as the tail and half of the body, and the body is as much as the head and the tail together.” How big is the fish? | 3. Let $2 x$ kg be the weight of the torso, then the head will weigh $x+1$ kg. From the condition that the torso weighs as much as the head and tail together, we get the equation: $2 x=x+1+1$. From this, $x=2$, and the whole fish weighs - 8 kg. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. The Clock Hands Problem. At what time after 12:00 will the minute hand first catch up with the hour hand? | 11. At 13:00, the minute hand will lag behind the hour hand by 5 minute divisions. Before the "meeting," the hour hand will travel $x$ divisions, and the minute hand will travel $12x$ divisions. From the equation $x + 5 = 12x$, we get that $x = \frac{5}{11}$ minute divisions on the clock face. Answer: 1 hour $5 \frac{5... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Divide equally. There is milk in an eight-liter bucket. How can you measure out 4 liters of milk using a five-liter bucket and a three-liter jar? | 4. It is convenient to write the solution in the form of a table:
| 8-liter can | 8 | 3 | 3 | 6 | 6 | 1 | 1 | 4 |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 5-liter can | 0 | 5 | 2 | 2 | 0 | 5 | 4 | 4 |
| 3-liter jar | 0 | 0 | 3 | 0 | 2 | 2 | 3 | 0 | | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In a family, there are six children. Five of them are respectively 2, 6, 8, 12, and 14 years older than the youngest, and the age of each child is a prime number. How old is the youngest? | 4. Suppose the child's age does not exceed 35 years. Let's list all the prime numbers: $2,3,5,7,11,13,17,19,23$, 29,31. It is clear that the age of the younger child is an odd number. The numbers 29 and 31 also do not fit. Let's find the age of the younger child. He cannot be 1 year old, because $1+8=9$. His age cannot... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13. In the senate of the country of knights and liars - there are 100 senators. Each of them is either a knight or a liar. It is known that: 1. At least one of the senators is a knight. 2. Out of any two arbitrarily chosen senators, at least one is a liar. Determine how many knights and how many liars are in the senate... | 13. Only one of the senators is a knight. The fact that there is at least one knight among the senators follows from the first statement, the existence of a second knight contradicts the second statement. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Scientific organization of labor. There are logs of two types: 6 meters and 7 meters long. They need to be sawn into 1-meter logs. Which logs are more profitable to saw? | 1. To saw 42 one-meter logs from six-meter logs requires 35 cuts, while from seven-meter logs - 36. It can be considered that sawing six-meter logs is more profitable. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. To the Mechanics and Mathematics Faculty! Several identical books and identical albums were bought. The books cost 10 rubles 56 kopecks. How many books were bought if the price of one book is more than one ruble higher than the price of an album, and 6 more books were bought than albums? | 5. Since each book is more expensive than a ruble, no more than 10 books were bought. Moreover, it is clear that no fewer than 7 books were bought (since at least one album was bought). The number 1056 is divisible by 8 and not divisible by $7,9,10$. Therefore, 8 books were bought. (MSU, Mechanics and Mathematics, 1968... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. One or two? Let's take all natural numbers from 1 to 1000000 and for each of them, calculate the sum of its digits. For all the resulting numbers, we will again find the sum of their digits. We will continue this process until all the resulting numbers are single-digit. Among the million resulting numbers, 1 and 2 w... | 8. Let's use the statement: if a number when divided by 9 has a remainder of \( d \), then the sum of its digits will have the same remainder. Which numbers from 1 to 1000000 are more: those that have a remainder of 1 when divided by 9, or those that have a remainder of 2? In the range from 1 to 999999, there are an eq... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. Two detectives met. Here is their dialogue:
- Do you have two sons?
- Yes, they are young, they don't go to school.
- By the way, the product of their ages equals the number of pigeons near us.
- That's not enough information.
- And I named the older one after you.
- Now I know how old they are.
How old are the so... | 7. The ages of the sons could have been equal, hence the product of their ages - a perfect square. Upon receiving information about the product of the ages, the second detective could not answer the question, therefore, the product can be factored in two ways. Answer: 1 and 4 years. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. In a chess tournament, 8 players participated. They scored $7, 6, 4, 4, 3, 2, 1, 5$ and 0.5 points respectively. How many points did the players who took the first four places lose in matches against the others? | 7. From the condition of the problem, it follows that the tournament winner and the participant who took second place did not lose a single point in their matches against the others. Further, the participants who shared third and fourth places could have scored 8 points against the rest, but scored 7 (they played one p... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11. Homework. Fedya was supposed to divide a certain number by 4 and add 15 to it, but Fedya multiplied this number by 4 and subtracted 15, yet he still got the correct answer. What was this number? | 11. Solving the equation: $0.25 x + 15 = 4 x - 15$, we get the answer: 8. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Linear function. The distance between villages $\mathcal{A}$ and $B$ is 3 km. In village $\mathcal{A}-300$ students, and in village $B-$ 200 students. Where should a school be built to minimize the total distance traveled by students on their way to school? | 4. It is clear that the school should be built on the segment $A B$, but where exactly? Let the distance from village $A$ to the school be $x$, then the total distance traveled by all schoolchildren on the way
 and six half-liter bottles (worth 15 kopecks each). In the store, there was milk sold by the liter for 22 kopecks. What is the maximum amount of milk he could bring home? He had no... | 11. By returning six half-liter bottles and one liter bottle, Anton will receive 1 ruble 10 kopecks, which will be the cost of 5 liters of milk. The 5 liters of milk he buys can be carried home in the remaining liter bottles. Let's ensure that he won't be able to carry more than 5 liters. If he returns not one liter bo... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
12. Given a 1998-digit number, every two adjacent digits of which form a two-digit number divisible by 17 or 23. The last digit of the number is 1. What is the first? | 12. Let's start recording this number from the end. At some point, we will notice that the number has the form ...92346...9234692346851. Further, we see that the digits 92346 are repeating, so we subtract from 1998 the number of digits that do not belong to this cycle - 3. We divide the resulting number by 5 (the numbe... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13. The warehouse has nails in boxes weighing $24, 23, 17$ and 16 kg. Can the warehouse keeper issue 100 kg of nails from the warehouse without opening the boxes? | 13. For example: 4 boxes - at 17 kg each and 2 boxes - at 16 kg each. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
30. What digit does the number $3^{100}$ end with? | 30. The number $3^{100}=81^{25}$, and therefore, ends in 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
31. What is the remainder when the number $2^{99}$ is divided by 7?
95 More problems! | 31. Note that $2^{99}=8^{33}=(7+1)^{33}=(7+1) \ldots(7+1)$. Expand the brackets. The resulting terms will be divisible by 7, except for 1. Thus, $2^{99}$ can be written in the form: $7 x$ +1 . Answer: 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
39. Calculate: $2379 \cdot 23782378-2378 \cdot 23792379$. | 39. Let $a=2378$, then the desired expression is: $(a+1)(10000 a+a)-a(10000(a+1)+(a+1))=$ $=(a+1) \cdot 10001 a-a \cdot 10001 \cdot(a+1)=0$ | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
41.On the farmyard, geese and piglets were wandering around. A boy counted the number of heads, there were 30, then he counted the total number of legs, there were 84. Can you find out how many geese and how many piglets were on the farmyard? | 41.If only geese were wandering around the farmyard, there would be 60 legs in total, the "extra" legs, which number 24, belong to the piglets - two for each. Therefore, there were 12 piglets, and 18 geese. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
42.A brick weighs 2 kg and half a brick. How much does the brick weigh? | 42.It follows from the condition that half a brick weighs 2 kg. Therefore, a whole brick weighs 4 kg. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
47. 6 carps are heavier than 10 perches, but lighter than 5 pikes; 10 carps are heavier than 8 pikes. What is heavier: 2 carps or 3 perches? | 47. Since 6 carp are heavier than 10 perch, it is clear that 6 carp are even heavier than 9 perch. Therefore, 2 carp are heavier than 3 perch. This means that two of the three conditions in the problem are redundant. | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
50.What is the 1997th digit in the decimal expansion of the fraction $\frac{1}{7}=0.142857 \ldots ?$ | 50.If we divide 1 by 7 using long division, we get that $\frac{1}{7}=0.(142857)$. The remainder of 1997 divided by 6 is 5, Therefore, the digit at the 1997th place is 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
51. Little One eats a jar of jam in six minutes, while Karlson is twice as fast. How long will it take them to eat the jam together | 51. The question of the problem can also be formulated as follows: "How long would it take for three Little Ones to eat the jam?" (According to the condition of the problem, Carlsson can be equated to two Little Ones). It is clear that three Little Ones would finish the jam three times faster than one. Answer: in 2 min... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
60.Yura left the house for school 5 minutes later than Lena, but walked at twice her speed. How long after leaving will Yura catch up to Lena? | 60.Let Lena walk $s$ km in 5 minutes. Then in the next 5 minutes, Yura will walk $2 s$ km, and Lena will walk another $s$ km, that is, a total of $2 s$ km. Therefore, in 5 minutes, Yura will catch up with Lena. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
61.If a cyclist rides at a speed of 10 km/h, he will be 1 hour late. If he rides at a speed of 15 km/h, he will arrive 1 hour early. At what speed should he ride to arrive on time? | 61.If there were two cyclists, with the first one's speed being $10 \mathrm{km} / \mathrm{h}$ and the second one's speed being $15 \mathrm{km} / \mathrm{h}$. Then, according to the problem, if the first cyclist started 2 hours earlier than the second, they would arrive at the destination simultaneously. In this case, i... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
64.A motorboat travels 90 km downstream in the same time it takes to travel 70 km upstream. What distance can a raft drift in the same time? | 64.Let in $t$ hours the motorboat covers 90 km downstream and 70 km upstream. Then the speed of the motorboat downstream is $\frac{90}{t}$ km/hour, and the speed upstream is $\frac{70}{t}$ km/hour. From this, the doubled speed of the current will be $\frac{90}{t}-\frac{70}{t}=\frac{20}{t}$ km/hour. The speed of the cur... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
69. Someone has 12 pints of wine and wants to pour out half, but he does not have a 6-pint container. However, he has two containers with capacities of 5 and 8 pints. How can he measure exactly 6 pints of wine? | 69. The solution is visible from the table:
| steps: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 12 l | 12 | 4 | 4 | 9 | 9 | 1 | 1 | 6 |
| 8 l | 0 | 8 | 3 | 3 | 0 | 8 | 6 | 6 |
| 5 l | 0 | 0 | 5 | 0 | 3 | 3 | 5 | 0 | | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
76. In the box, there are 100 white, 100 red, 100 blue, and 100 black balls. What is the smallest number of balls that need to be pulled out, without looking into the box, to ensure that among them there are at least 3 balls of the same color? | 76. Answer: 9 balls. If we randomly draw 8 balls, there might not be three balls of the same color (2 white + 2 red + 2 blue + 2 black). If we add one more ball, then there will definitely be 3 balls of the same color. | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
78. How many circles of radius 1 are needed to cover a square with a side length of $2$? | 78. A circle of radius 1 can only cover one vertex of a $2 \times 2$ square, so at least four circles are required. However, a circle of unit radius can completely cover a $1 \times 1$ square, so four circles are sufficient. Answer: 4 circles. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
81. In the box, there are pencils: 7 red and 5 blue. In the dark, pencils are taken. How many pencils need to be taken to ensure that there are at least two red and at least three blue among them? | 81. In order to definitely take no less than two red pencils, you need to take no less than 7 pencils, and to definitely take no less than 3 blue ones, you need to take no less than 10 pencils. Therefore, to definitely take no less than two red and no less than 3 blue, you need to take no less than 10 pencils. | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11. Find the maximum value of the function $y=3 x+4 \sqrt{1-x^{2}}$
$$ | \text { S o l u t i o n. }
$$
First, let's find the domain of the given function: $D(f)=$ $=\left\{x \mid 1-x^{2} \geqslant 0\right\}=\{x \mid-1 \leqslant x \leqslant 1\} ; \quad E(f)=\{y \mid \quad$ the equation $y=3 x+4 \sqrt{1-x^{2}}$ has a solution on $\left.[-1 ; 1]\right\}=\{y \mid$ the equation $y-3 x=4 \sqrt{1... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 35. For the fattening of animals, two types of feed I and II are used. Each kilogram of feed I contains 5 units of nutrient $\boldsymbol{A}$ and 2.5 units of nutrient B, while each kilogram of feed II contains 3 units of nutrient A and 3 units of nutrient B. Experimental data has shown that the fattening of ani... | ## S o l u t i o n.
1. Let's construct a mathematical model of this problem. Let $x$ and $y$ be the number of kilograms of feed of types I and II, respectively, consumed daily. Then the system of constraints is:
$$
\left\{\begin{array}{l}
5 x+3 y \geqslant 30 \quad(a) \\
2.5 x+3 y \geqslant 22.5 \\
x \geqslant 0, y \... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 37. On the set of solutions of the system of constraints
$$
\left\{\begin{array}{l}
2-2 x_{1}-x_{2} \geqslant 0 \\
2-x_{1}+x_{2} \geqslant 0 \\
5-x_{1}-x_{2} \geqslant 0 \\
x_{1} \geqslant 0, \quad x_{2} \geqslant 0
\end{array}\right.
$$
find the minimum value of the function $F=x_{2}-x_{1}$. | S o l u t i o n.
The set of feasible plans is the polygon $A B C D E$ (Fig. 35). The line $2-$ $-x_{1}-x_{2}=0$ is parallel to the level lines of the function $F=x_{2}$ $-x_{1}$. Therefore, all points on the segment $C D$ give the same minimum value of the objective function $F=x_{2}-x_{1}$ on the set of points of the... | -2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Problem 39. Minimize $\boldsymbol{F}=\boldsymbol{x}_{2}-\boldsymbol{x}_{1}$ for non-negative $x_{1}$ and $x_{2}$, satisfying the system of constraints:
$$
\left\{\begin{aligned}
-2 x_{1}+x_{2}+x_{3} & =2 \\
x_{1}-2 x_{2}+x_{4} & =2 \\
x_{1}+x_{2}+x_{5} & =5
\end{aligned}\right.
$$
56 | ## S o l u t i o n.
These constraints can be considered as derived from inequalities, since each of the variables $x_{3}, x_{4}, x_{5}$ appears only in one equation.
1. Write the constraints as equations expressing the basic variables in terms of the non-basic variables:
$$
\left\{\begin{array}{l}
x_{3}=2+2 x_{1}-x_... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. If a zero is appended to the right of the thought number and the result is subtracted from 143, the result will be three times the thought number. What number was thought of? | Solution. Let the number be $x$. We have the equation $143-10 x=3 x$, from which $x=11$. A n s w e r: 11 . | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
41(1283). The path from $A$ to $B$ goes 3 km uphill, 6 km downhill, and 12 km on flat ground. The motorcyclist covered this path in 1 hour and 7 minutes, and the return path in 1 hour and 16 minutes. Find the motorcyclist's speed uphill and downhill, if his speed on flat ground was 18 km/h. (Note that the motorcyclist ... | Solution. I method. On a flat surface in one direction, the motorcyclist traveled $\frac{2}{3}$ hours $\left(12: 18=\frac{2}{3}\right)$, or 40 minutes. Then, 3 km uphill and 6 km downhill, the motorcyclist traveled 27 minutes, and 6 km uphill and 3 km downhill, the motorcyclist traveled 36 minutes. If we denote the mot... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
45 (1287). A rider and a pedestrian simultaneously set off from point $A$ to point $B$. The rider, arriving in $B$ 50 minutes earlier than the pedestrian, returned back to $A$. On the return trip, he met the pedestrian 2 kilometers from $B$. The rider spent 1 hour and 40 minutes on the entire journey. Find the distance... | S o l u t i o n. Since the rider spent 1 hour 40 minutes on the entire journey, he spent 50 minutes on the journey from $A$ to $B$ (1 hour 40 minutes: $2=50$ minutes). Then the pedestrian spent 1 hour 40 minutes on the journey from $A$ to $B$, as he arrived in $B$ 50 minutes later than the rider. Therefore, the pedestr... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
47 (1289). Two brothers walk together from school to home at the same speed. One day, 15 minutes after leaving school, the first brother ran back to school and, upon reaching it, immediately set off to catch up with the second. Left alone, the second continued walking home at half his usual speed. When the first brothe... | Solution. Let the initial speed of the brothers be denoted by $v$ (m/min), and the segment of the path that the second brother traveled at a speed of $\frac{v}{2}$ be denoted by $s$. For this segment, he spent $\frac{2 s}{v}$ minutes $\left(s: \frac{v}{2}\right)$, which is 6 minutes more than the usual time spent (at s... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
105(974). Determine the value of $b$, if the graph of the function defined by the formula passes through the point with coordinates $(3 ; 10)$:
1) $y=x+b$;
2) $y=3 x+b$
3) $y=-\frac{1}{3} x+b$;
4) $y=-\frac{1}{2} x+b$.
A n s w e r: 1) $b=7$; 2) $b=1$; 3) $b=11$; 4) $b=11.5$.
106(975). Define the function by a formul... | Solution. Let $x$ be the number of turns in the spring. Considering that the number of gaps in the spring is one less than the number of turns, according to the problem, we set up the equation $5x + 8(x-1) = 122$, from which $x = 10$.
Answer: 10 turns. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
108(977). A person, living in a cottage near a railway station, usually manages to walk from the cottage to the station in time for the train departure in 18 minutes. Once, this person was delayed at home before leaving by several minutes. Although after that, he walked 1.2 times faster than usual, he still missed the ... | Solution. If the usual speed of the gardener is $v$ m/min, then the distance from the garden to the station is $18 v$ meters. When the gardener walked at a speed of $1.2 v$ m/min, it took 15 minutes to travel from home to the garden $(18 v: 1.2 v=15)$. Since the gardener was 2 minutes late, the train left 13 minutes af... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
120 (989). The road from point $A$ to point $B$ is 11.5 km long and goes first uphill, then on flat ground, and finally downhill. The pedestrian spent 2 hours and 54 minutes on the journey from $A$ to $B$, and 3 hours and 6 minutes on the return journey. The pedestrian's walking speed uphill was 3 km/h, on flat ground ... | Solution. Let the length of the road uphill be $x$ kilometers, on the plain be $y$ kilometers, and downhill be $z$ kilometers. According to the problem, we form a system of three linear equations with three unknowns:
88
$$
\left\{\begin{array}{l}
\frac{x}{3}+\frac{y}{4}+\frac{z}{5}=2.9 \\
\frac{z}{3}+\frac{y}{4}+\fra... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
139. On a meadow, grass grows. If 9 cows were let onto the meadow, they would empty it in 4 days. If 8 cows were let onto the meadow, they would eat all the grass in 6 days. How many cows can graze on the meadow all the time while the grass grows? | Solution. Let $x$ kilograms of grass grow on the meadow per day, and one cow eats $y$ kilograms of grass per day. We have the system of equations:
$$
\left\{\begin{array}{l}
a+4 x-9 \cdot 4 y=0 \\
a+6 x-8 \cdot 6 y=0
\end{array}\right.
$$
where $a$ is the amount of grass that grew on the meadow before the cows were l... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
158. It is known that the number $a$ is $n$ times greater than the number $b$, and the sum of the numbers $a$ and $b$ is $m$ times greater than their difference. Find the sum of the numbers $m$ and $n$, if $m$ and $n$ are natural numbers.
94 | Given the problem, we set up the system of equations:
$$
\left\{\begin{array}{l}
\frac{a}{b}=n \\
\frac{a+b}{a-b}=m
\end{array}\right.
$$
From the second equation, express $\frac{a}{b}$. We have:
$$
\begin{gathered}
a+b=m a-m b, a(1-m)=-b(m+1) \\
\frac{a}{b}=\frac{m+1}{m-1}=\frac{m-1+2}{m-1}=1+\frac{2}{m-1} .
\end{g... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
173. In the Olympiad, 55 schoolchildren participated. All of them submitted their work. During the checking of each problem, one of three grades was given: «+» - the problem was solved, «-» - the problem was attempted, «0» - the problem was not attempted. After checking all the works, it turned out that in no two works... | Instruction. If the number of tasks was $a$, then
$$
\frac{(a+1)(a+2)}{2}=55 \text { when } a=9 .
$$
A n s w e r: 9 tasks. | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
15(1091). Prove that the value of the expression $\sqrt{11+6} \sqrt{ } 2+$ $+\sqrt{11-6 \sqrt{2}}$ is a natural number. | Proof. I method. Let
$$
\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}}=A
$$
Then
$$
A^{2}=11+6 \sqrt{2}+11-6 \sqrt{2}+2 \sqrt{121-72}=36
$$
from which
$$
\sqrt{A^{2}}=|A|=6, \text { i.e. }|\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}}|=6
$$
Since under the modulus sign is the sum of two radicals, each of which is a positi... | 6 | Algebra | proof | Yes | Yes | olympiads | false |
23(1099). A bus left city $M$ for city $N$ at a speed of 40 km/h. A quarter of an hour later, it met a car traveling from city $N$. This car reached city $M$, 15 minutes later it set off back to city $N$ and overtook the bus 20 km from city $N$. Find the distance between cities $M$ and $N$, if the speed of the car is 5... | Solution. Since the bus traveled 10 km in a quarter of an hour $(40: 4=10)$, the bus met the car 10 km from city $M$. If the distance between the cities is $s$ kilometers, then by the time the car overtook the bus, the bus had traveled $(s-30)$ kilometers from the moment of the meeting, spending $\frac{s-30}{40}$ hours... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
33(1109). Find the members of the proportion $x_{1}: x_{2}=x_{3}: x_{4}$, where the first term is 6 more than the second, and the third is 5 more than the fourth. The sum of the squares of all terms is 793. | Solution. By the condition $x_{1}=x_{2}+6, x_{3}=x_{4}+5,\left(x_{2}+6\right): x_{2}=$ $=\left(x_{4}+5\right): x_{4}$.
We have a system of two equations with two unknowns:
$$
\left\{\begin{array}{l}
\left(x_{2}+6\right)^{2}+x_{2}^{2}+\left(x_{4}+5\right)^{2}+x_{4}^{2}=793 \\
\left(x_{2}+6\right) \cdot x_{4}=x_{2} \cd... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
46(1122). Several points are marked on a plane, no three of which lie on the same line. A line is drawn through every two points. How many points are marked on the plane if it is known that a total of 45 lines have been drawn? | Solution. The problem reduces to solving the equation $\frac{x(x-1)}{2}=45$ in natural numbers, where $x$ is the number of points marked on the plane. We have $x^{2}-x-90=0$, from which $x_{1}=10, x_{2}=-9$. Only the root $x=10$ satisfies the condition of the problem.
Answer: 10 points.
$\triangle 47(969)$ '. Prove t... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
51 (974). What is the last digit of the difference $43^{43}-17^{17}$? | Solution. Let's determine the last digits of various powers of the numbers 43 and 17. The number 43 ends with the digit $3, 43^2$ ends with the digit $9, 43^3$ ends with the digit $7, 43^4$ ends with the digit 1. Therefore, the last digits in subsequent powers of the number 43 will repeat: $3, 9, 7, 1$, and so on. Sinc... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
55 (978). There are 4 balls of different masses. Using a balance scale without weights, how many weighings are needed to arrange these balls in order of decreasing mass? | Solution. Let $m_{A}, m_{B}, m_{C}, m_{D}$ be the masses of four different balls. We will use a graph to show how the weighings are performed (the arrow points from the heavier ball to the lighter one).
Suppose, for example, after two weighings, we have $m_{A}>m_{B}$ and $m_{C}>m_{D}$ (Fig. $29, a$ ). Next, we will co... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
69(1003). A cyclist and a motorcyclist are moving towards an intersection along two mutually perpendicular roads. At a certain moment, the cyclist is 8 km away, and the motorcyclist is 15 km away from the intersection. After how many minutes will the distance between them be 5 km, if the cyclist's speed is $\frac{1}{3}... | Solution. Let $OA=8$ km, $OB=15$ km (Fig. 30). If the distance of 5 km between the motorcyclist and the cyclist will be in $t$ minutes $\left(A_{1} B_{1}=5\right.$ km), then $A A_{1}=\frac{1}{3} t$ (km), $B B_{1}=t$ (km). Then $O A_{1}=O A-\frac{1}{3} t=$
. A master and his apprentice were supposed to complete a job by a certain deadline. However, when half of the work was done, the apprentice fell ill, and the master, left alone, finished the job 2 days late. How many days would it take for each of them to complete the entire job working alone, if the master wo... | Solution. Let the master be able to complete the entire job in $x$ days, then the apprentice - in $(x+5)$ days. In one day, the master completed $\frac{1}{x}$ of the work, the apprentice completed $\frac{1}{x+5}$ of the work; when working together, the master and the apprentice completed $\frac{2 x+5}{x(x+5)}$ of the w... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
84. Nine identical books cost 11 rubles and some kopecks, while 13 such books cost 15 rubles and some kopecks. Determine the exact cost of one book | S o l u t i o n. It is clear that each book costs 1 r. $m$ k., where $m$ is an integer, and $m<100$. From the first condition of the problem, it follows that $200<9 m<300$, i.e., $23 \leqslant m \leqslant 33$. From the second condition, $200<13 m<300$, i.e., $16 \leqslant m \leqslant 23$. Therefore, $m=23$.
A n s w e ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
89. Calculate the value of the expression $\frac{2 a-b}{3 a-b}+\frac{5 b-a}{3 a+b}$, given that $10 a^{2}-3 b^{2}+5 a b=0$ and $9 a^{2}-b^{2} \neq 0$. | Solution. Since $5 a b=3 b^{2}-10 a^{2}$, then
$$
\begin{aligned}
& \quad \frac{2 a-b}{3 a-b}+\frac{5 b-a}{3 a+b}=\frac{3 a^{2}+15 a b-6 b^{2}}{9 a^{2}-b^{2}}=\frac{3 a^{2}+3\left(3 b^{2}-10 a^{2}\right)-6 b^{2}}{9 a^{2}-b^{2}}= \\
& \text { Answer: }-3 . \\
& =\frac{-3\left(9 a^{2}-b^{2}\right)}{9 a^{2}-b^{2}}=-3 .
\... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
96. The number 392 was divided by a natural number $a$, and from the quotient, $a$ was subtracted. The same operation was performed with the resulting difference, and the same operation was performed again with the new result. The final answer was the number $-a$. What is the value of $a$? | Solution. According to the problem, we form the equation
$$
\left(\left(\frac{392}{a}-a\right): a-a\right): a-a=-a
$$
from which
$$
\begin{aligned}
\left(\frac{392}{a}-a\right): a-a & =0, \\
a^{3}+a^{2}-392=0, a^{2}(a+1) & =392, \text { where } a \in \boldsymbol{N} .
\end{aligned}
$$
We have that 392 is divisible b... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
101. Given two quadratic equations $x^{2}-x+m=0, x^{2}-x+$ $+3 m=0, m \neq 0$. Find the value of $m$ for which one of the roots of the second equation is twice the root of the first equation. | Solution. Let $x_{0}$ be the root of the first equation, and $2 x_{0}$ be the root of the second. Then the following numerical equalities hold:
$$
\begin{gathered}
x_{0}^{2}-x_{0}+m=0 \\
4 x_{0}^{2}-2 x_{0}+3 m=0
\end{gathered}
$$
Subtracting the second equality from three times the first, we get $-x_{0}^{2}-x_{0}=0$... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
106. Several chess players participated in a match tournament, where each participant played several games against every other participant. How many rounds did this competition consist of, if a total of 224 games were played? | Solution. Let $x$ be the number of participants in the competition, and $y$ be the number of rounds. In one round, each chess player plays $(x-1)$ games, and all participants together play $\frac{x(x-1)}{2}$ games. Therefore, the total number of games played in the competition is $\frac{x(x-1) y}{2}$. Consequently, $x(... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
115. Find the minimum value of the expression $\frac{(4+x)(1+x)}{x}$, where $x$ is a positive number. | Solution. Since
$$
\frac{(4+x)(1+x)}{x}=\frac{4}{x}+x+5=2\left(\frac{2}{x}+\frac{x}{2}\right)+5
$$
and $\frac{2}{x}+\frac{x}{2} \geqslant 2$ (for $c>0$, the inequality $c+\frac{1}{c} \geqslant 2$ holds, i.e., 2 is the minimum value of the sum of two reciprocals (see 627 from [2])), the expression $2\left(\frac{2}{x}+... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
116. Find the minimum value of the fraction $\frac{x^{2}-3 x+3}{1-x}$, if $x<1$. | Solution. I method. The given fraction can be represented as $\frac{x^{2}}{1-x}+3$. For $x<1$, the first term is non-negative and equals 0 only when $x=0$. Since 0 is included in the set of values for $x$, the given fraction achieves its minimum value of 3 when $x=0$.
II method. Representing the given fraction as the ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
117. Find the minimum value of the expression
$$
(x-5)(x-1)(x-6)(x-2)+9
$$ | S o l u t i o n. We have:
$$
\begin{gathered}
y=(x-5)(x-1)(x-6)(x-2)+9=\left(x^{2}-7 x+10\right) \times \\
\times\left(x^{2}-7 x+6\right)+9=\left(x^{2}-7 x\right)^{2}+16\left(x^{2}-7 x\right)+69= \\
=\left(x^{2}-7 x+8\right)^{2}+5
\end{gathered}
$$
From here, $y_{\text {min }}=5$, when $x^{2}-7 x+8=0$, i.e., when
$$... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
121. Find the value of the difference
$$
A=\sqrt{|12 \sqrt{5}-29|}-\sqrt{12 \sqrt{5}+29}
$$ | S o l u t i o n. Noting that $|12 \sqrt{5}-29|=29-12 \sqrt{5}$, we get:
$$
A=\sqrt{29-12 \sqrt{5}}-\sqrt{29+12 \sqrt{5}}
$$
Since
$$
29-12 \sqrt{5}=20-12 \sqrt{5}+9=(2 \sqrt{5}-3)^{2}
$$
and $29+12 \sqrt{5}=(2 \sqrt{5}+3)^{2}$, we obtain that
$$
A=2 \sqrt{5}-3-(2 \sqrt{5}+3)=-6
$$
A n s w e r: -6 . | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
123. How many right-angled triangles are there with sides of integer lengths, if one of the legs of these triangles is equal to 15? | Solution. If $x$ is the hypotenuse and $y$ is the unknown leg of a right triangle, then $x^{2}-y^{2}=15^{2} ;(x-y)(x+y)=$ $=3 \cdot 3 \cdot 5 \cdot 5$
Since $(x-y)$ and $(x+y)$ are natural numbers, and $x+y > x-y$, there are only 4 cases:
$$
\left\{\begin{array} { l }
{ x - y = 1 } \\
{ x + y = 225 }
\end{array} \le... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
130. Find the smallest value of the expression $\left|36^{m}-5^{n}\right|$, where $m$ and $n$ are natural numbers. | Solution. The difference $36^{m}-5^{n}$ ends with the digit 1, and the difference $5^{n}-36^{m}$ ends with the digit 9. We will prove that $\left|36^{m}-5^{n}\right|$ cannot be equal to either 1 or 9.
From the equation $36^{m}-5^{n}=1$, it follows that $\left(6^{m}-1\right)\left(6^{m}+1\right)=5^{n}$, and then the pow... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
131. A natural number $n$ is the product of four consecutive natural numbers, each greater than 5. Determine the maximum number of possible last digits of $n$, given that its last digit is not 0. | Solution. Among four consecutive numbers whose product is equal to $n$, there are two even numbers, and therefore the number $n$ is even. Consequently, none of these numbers is divisible by 5 - otherwise, the number $n$ would end in the digit 0. Thus, we have:
$$
\begin{gathered}
n=(5 k+1)(5 k+2)(5 k+3)(5 k+4)=\left(2... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
139. Can you measure out 10 liters of water using two containers of 9 and 11 liters? | Solution. Fill the 11-liter vessel (vessel $A$) with water, then use this water to fill the 9-liter vessel (vessel $B$), empty $B$, and transfer the remaining 2 liters from $A$ to $B$. After this, vessel $A$ will be empty, and vessel $B$ will have 2 liters. Repeating these operations three more times, we will end up wi... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
140. Find the greatest common divisor of all nine-digit numbers composed of the digits $1,2,3,4,5,6,7,8,9$ (without repetition). | Solution. Each of the 9! such numbers is divisible by 9, since the sum of the digits of each number, which is 45, is divisible by 9. We will prove that 9 is the greatest common divisor of these numbers.
Let the greatest common divisor of the considered numbers be greater than 9. Denote it by $d$. Then the difference b... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
141. Find the greatest common divisor of all six-digit numbers composed of the digits $1,2,3,4,5,6$ (without repetition). | Re shenie. Each of the 6! numbers is divisible by 3 (since the sum of the digits is 21). We will show that 3 is the greatest common divisor of these numbers.
Using the divisibility rules of numbers, it is easy to show that the greatest common divisor of the considered six-digit numbers cannot be equal to $2, 4, 5, 6, ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
144(1172). Two athletes are running on the same closed track. The speed of each is constant, and one takes 5 seconds less than the other to run the entire track. If they start running simultaneously from the same starting line in the same direction, they will be side by side after 30 seconds. How long will it take for ... | Solution. Let $s$ (m) be the length of the closed track, $t$ (s) the time it takes for the first athlete to run the entire track. Then the speed of the first athlete is $\frac{s}{t}(\mathrm{~m} / \mathrm{s})$, and the speed of the second athlete is $\frac{s}{t+5}(\mathrm{m} / \mathrm{s})$.
According to the problem, we... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
145(1173). A pedestrian and a cyclist set off from point $A$ to point $B$ simultaneously. At point $B$, the cyclist turns back and meets the pedestrian 20 minutes after the start of the journey. Without stopping, the cyclist continues to point $A$, turns back, and catches up with the pedestrian 10 minutes after the fir... | Solution. Let the speed of the pedestrian be $v_{1}$ meters per minute, the speed of the cyclist be $v_{2}$ meters per minute, and $t$ minutes be the time the pedestrian spent from the moment of the second meeting with the cyclist until arriving at point $B$.
According to the problem, we form a system of equations:
$... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
146(1174). A motorcyclist left point $A$ for point $B$ and at the same time a cyclist left point $B$ for point $A$. The motorcyclist arrived at point $B$ 2 hours after meeting the cyclist, while the cyclist arrived at point $A$ 4.5 hours after meeting the motorcyclist. How many hours were the motorcyclist and the cycli... | Solution. Let the motorcyclist and the cyclist have been on the road for $t$ hours before meeting. If $v_{1}$ kilometers per hour is the speed of the motorcyclist, and $v_{2}$ kilometers per hour is the speed of the cyclist, then we have the system of equations:
$$
\left\{\begin{array}{l}
v_{1}=\frac{v_{2} t}{2} \\
v_... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4(1132). For what values of $a$ do the quadratic trinomials $x^{2}+$ $+a x+1$ and $x^{2}+x+a$ have a common root? | Solution. Let $x=x_{1}$ be the common root of the given quadratic trinomials. Then $x_{1}^{2}+a x_{1}+1=0$ and $x_{1}^{2}+x_{1}+a=0$, i.e., $x_{1}^{2}+a x_{1}+1=x_{1}^{2}+x_{1}+a, \quad a \cdot\left(x_{1}-1\right)=x_{1}-1, \quad\left(x_{1}-1\right) \cdot(a-1)=0$. From this, $a=1$ or $x_{1}=1$.
When $a=1$, each of the ... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5 (1133). For what value of $a$ does the sum of the squares of the roots of the quadratic trinomial $x^{2}-(a-2) x-a-1$ take the smallest value? | Solution. Based on Vieta's theorem, we have $x_{1}+x_{2}=$ $=a-2, x_{1} \cdot x_{2}=-a-1$. Then
$$
\begin{gathered}
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(a-2)^{2}-2(-a-1)= \\
=a^{2}-2 a+6=(a-1)^{2}+5
\end{gathered}
$$
Since $(a-1)^{2} \geqslant 0$, $x_{1}^{2}+x_{2}^{2}$ takes its minimum val... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.